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Make s the subject of: v^2= u^2+2as

The aim of this question is to algebraically manipulate the equation to end up with an equation with s on its own, on one side, making it the subject of the equation.We will start by looking at the terms in ...
DD
Answered by Divya D. • Maths tutor
66374 Views

Prove algebraically that 0.256565656... can be written as 127/495

Let x = 0.2565656...Then 10x = 2.565656.... and 1000x = 256.565656...So 1000x - 10x = 256.565656... - 2.565656... = 254990x = 254 x = 254/990 = 127/495
OT
Answered by Oliver T. • Maths tutor
16439 Views

Solve the simultaneous equations algebraically.

x^2 + y^2 = 29y - x = 3Rearrange the second equation such that one variable, either x or y, is the subject. I will rearrange to make x the subject. x = y - 3. Substitute the new equation into equation 1 in p...
JA
Answered by Jane A. • Maths tutor
3575 Views

Prove that (3n+1)²-(3n-1)² is a multiple of 4 taking into account that n is a positive integer value

Square the brackets (3n+1)²= (3n+1)(3n+1) = 9n²+3n+3n+1 = 9n²+6n+1 (3n-1)²= (3n-1)(3n-1) = 9n²-3n-3n+1 = 9n²-6n+12. Write out the full equation (9n²+6n+1) - (9n²-6n+1) = 9n²+6n+1-9n²+6n-1 = 12n3. Explain you...
NK
Answered by Nalin K. • Maths tutor
7749 Views

How do you solve two simultaneous equations? (i.e. 5x + y =21 and x - 3y =9)

There are two ways of solving the simultaneous equations. The easiest one is to write two equations beneath each other and then try to get either x or y values the same on both of them by multiplying the val...
PM
Answered by Paulina M. • Maths tutor
3706 Views